niks18 wrote:
Mr. Bond has forgotten his \(n\) digit locker code. He remembers that the first three digits are either \(853\) or \(847\) and there was a number that appeared only once in the code. If Mr. Bond were to use a trial and error process to open his locker, what is the minimum number of trials he has to make before he can be certain to succeed?
OETo solve this Permutation problem we need three values-
a) the value of n = no of digit in the locker code
b) the digit that appeared only once
c) whether the locker code was ODD or EVEN because permutation will get impacted based on even/odd combination as we have one number that can be used only once
Statement 1:Quote:
Mr. Bond recalled that the number of digits in the code was a single digit prime number which, when multiplied by another prime number, returned an even number greater than \(10\)
this implies that \(n=7\) because \(7*2=14>10\)
Quote:
He also recalled that the digit that appeared only once was square of a single digit odd prime number
this implies that the digit that appeared only once was \(9\)=\(3^2\)
Quote:
and the locker code was not a multiple of two.
this implies that the code is ODD
Thus we got all the information we were looking for. Hence
Sufficient. Note: this is a DS problem hence we don't need an exact answer here.
Statement 2:Quote:
The locker code is odd, and the number nine appears once in the code. \(n<14\) and \(n=6k+1\), where \(k\) is any integer.
From this we get the information that the code is ODD and
9 appears only once, but the value of \(n\) cannot be determined,
if \(k=1\), \(n=7\) & if \(k=2\), \(n=13\). Thus two values of n are possible. Hence
InsufficientOption
A---------------------------------------------------------------------------------------------------------------------------
FYI - calculation for number of trials
There are two possible cases. The number \(9\) comes at the end, or it comes at position 4th, 5th, or 6th.
For the first case, the number would look like: \(853 - - - 9\) or \(847 - - - 9\)
In both these cases, the blanks can be occupied by any of the available \(9\) digits (0, 1, 2, ..., 8).
Thus, total possible numbers would be \(2 × (9 × 9 × 9) = 1458\).
For the second case, the number \(9\) can occupy any of the given position 4th, 5th, or 6th, and there shall be an odd number at position 7th.
Thus, the total number of ways shall be \(2[3(9 × 9 × 4)] = 1944\). Hence, answer is \(1458+1944=3402\).